Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long int pre[200003];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int sy, a, b;
  cin >> sy >> a >> b;
  long long int n;
  cin >> n;
  vector<array<long long int, 2>> v;
  for (long long int i = 0; i < n; i++) {
    long long int l, r;
    cin >> l >> r;
    v.push_back({l, r});
  }
  for (long long int i = 0; i < n; i++) {
    pre[i + 1ll] = pre[i] + v[i][1] - v[i][0];
  }
  long long int q;
  cin >> q;
  double ans;
  while (q--) {
    long long int x, y;
    cin >> x >> y;
    double xs, xe, m;
    long long int indxs, indxe;
    if (x == a) {
      xs = x;
    } else {
      m = (y - sy) / (double)(x - a);
      xs = (m * x - y) / m;
    }
    long long int lo = 0;
    long long int hi = n - 1ll;
    long long int indx = -1ll;
    while (lo <= hi) {
      long long int mid = (lo + hi) / 2;
      if (v[mid][0] <= xs && xs <= v[mid][1]) {
        indx = mid;
        break;
      } else if (xs < v[mid][0]) {
        hi = mid - 1ll;
      } else {
        lo = mid + 1ll;
      }
    }
    if (indx == -1) {
      long long int lo = 0;
      long long int hi = n - 1ll;
      while (lo <= hi) {
        long long int mid = (lo + hi) / 2;
        if (v[mid][0] >= xs) {
          indx = mid;
          hi = mid - 1ll;
        } else {
          lo = mid + 1ll;
        }
      }
      if (indx != -1ll) {
        xs = v[indx][0];
        ans = v[indx][1] - v[indx][0];
        indxs = indx;
        double z;
        if (x == xs) {
          z = x;
        } else {
          m = y / (x - xs);
          z = (sy / m) + xs;
        }
        if (z > b) {
          ans = 0;
          cout << fixed << setprecision(10) << ans << "\n";
          continue;
        }
      } else {
        ans = 0;
        cout << fixed << setprecision(10) << ans << "\n";
        continue;
      }
    } else {
      indxs = indx;
      ans = v[indx][1] - xs;
    }
    if (x == b) {
      xe = x;
    } else {
      m = (y - sy) / (double)(x - b);
      xe = (m * x - y) / m;
    }
    lo = 0;
    hi = n - 1ll;
    indx = -1ll;
    while (lo <= hi) {
      long long int mid = (lo + hi) / 2;
      if (v[mid][0] <= xe && xe <= v[mid][1]) {
        indx = mid;
        break;
      } else if (xe < v[mid][0]) {
        hi = mid - 1ll;
      } else {
        lo = mid + 1ll;
      }
    }
    if (indx == -1) {
      lo = 0;
      hi = n - 1ll;
      indx = -1ll;
      while (lo <= hi) {
        long long int mid = (lo + hi) / 2;
        if (v[mid][1] <= xe) {
          indx = mid;
          lo = mid + 1ll;
        } else {
          hi = mid - 1ll;
        }
      }
      if (indx != -1ll) {
        xe = v[indx][1];
        indxe = indx;
        ans += v[indx][1] - v[indx][0];
        double z;
        if (x == xe) {
          z = x;
        } else {
          m = y / (x - xe);
          z = (sy / m) + xe;
        }
        if (z < a) {
          ans = 0;
          cout << fixed << setprecision(10) << ans << "\n";
          continue;
        }
      } else {
        ans = 0;
        cout << fixed << setprecision(10) << ans << "\n";
        continue;
      }
    } else {
      indxe = indx;
      ans += xe - v[indx][0];
    }
    if (indxs == indxe) {
      ans = xe - xs;
    } else {
      ans += pre[indxe] - pre[indxs + 1ll];
    }
    double mult = (y - sy) / (double)y;
    ans = ans * mult;
    cout << fixed << setprecision(10) << ans << "\n";
  }
}